Sin(П/2+t)- сos(п+t)+1=0

Предыдущий
вопрос Следующий
вопрос

To find the value of t that satisfies the equation sin(π/2 + t) - cos(π + t) + 1 = 0, we can rewrite the sine and cosine functions in terms of the angle t.

sin(π/2 + t) = cos(t) and cos(π + t) = -cos(t)

Substitute these values into the equation to get:

cos(t) - (-cos(t)) + 1 = 0
2cos(t) + 1 = 0
2cos(t) = -1
cos(t) = -1/2

Since the cosine function is negative in the second and third quadrants, we can find the reference angle by taking the inverse cosine of -1/2:

t = arccos(-1/2)
t = 2π/3 or t = 4π/3

Therefore, the values of t that satisfy the equation sin(π/2 + t) - cos(π + t) + 1 = 0 are t = 2π/3 and t = 4π/3.